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Outcome 2
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Higher MathsHigher Maths
www.mathsrevision.comwww.mathsrevision.com
What is a setFunction in various formats
Composite FunctionsExponential and Log Graphs
Connection between Radians and degrees & Exact valuesSolving Trig EquationsBasic Trig Identities
Graph TransformationsTrig Graphs
Inverse function
Mindmap
Exam Question Type
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Sets & FunctionsSets & Functions
Notation & Terminology
SETS: A set is a collection of items which have some common property.
These items are called the members or elements of the set.
Sets can be described or listed using “curly bracket” notation.
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Sets & FunctionsSets & Functions
N = {natural numbers}
= {1, 2, 3, 4, ……….}
W = {whole numbers} = {0, 1, 2, 3, ………..}Z = {integers} = {….-2, -1, 0, 1, 2, …..}
Q = {rational numbers}
This is the set of all numbers which can be written as fractions or ratios.
eg 5 = 5/1 -7 = -7/1 0.6 = 6/10 = 3/5
55% = 55/100 = 11/20 etc
We can describe numbers by the following sets:
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R = {real numbers}This is all possible numbers. If we plotted values on a number line then each of the previous sets would leave gaps but the set of real numbers would give us a solid line.
We should also note that
N “fits inside” W
W “fits inside” Z
Z “fits inside” Q
Q “fits inside” R
Sets & FunctionsSets & Functions
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Sets & FunctionsSets & Functions
QZWN
When one set can fit inside another we say
that it is a subset of the other.
The members of R which are not inside Q are called irrational numbers. These cannot be expressed as
fractions and include , 2, 35 etc
R
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To show that a particular element/number belongs to a particular set we use the symbol
. eg 3 W but 0.9 Z
Examples
{ x W: x < 5 }= { 0, 1, 2, 3, 4 }
{ x Z: x -6 } = { -6, -5, -4, -3, -2, …….. }
{ x R: x2 = -4 } = { } or
This set has no elements and is called the empty set.
Sets & FunctionsSets & Functions
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Nat 5
What are Functions ?
Functions describe how one quantity
relates to another
Car Part
s
Assembly line
Cars
Defn: A function or mapping is a relationship between two sets in which each member of the first set is connected to exactly one member in the
second set.
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Nat 5
What are Functions ?
Functions describe how one quantity
relates to another
Dirty
Washing Machine
Clean
OutputInputyx
Functionf(x)
y = f(x)
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Nat 5
Defining a Functions
A function can be thought of as the relationship between
Set A (INPUT - the x-coordinate)
and
SET B the y-coordinate (Output) .
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Functions & MappingsFunctions & Mappings
A function can be though of as a black box
x - Coordinate
Input
Domain
Members (x - axis)Co-Domain
Members (y - axis)
Image
Range
Function
Output
y - Coordinatef(x) = x2+ 3x - 1
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Nat 5
Finding the Function
Find the output or input values for the functions below :
6
7
8
36
49
64
f(x) = x2
f: 0
f: 1
f:2
-1
3
7f(x) = 4x - 1
4 12
f(x) = 3x
5 15
6 18
Examples
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Functions & MappingFunctions & Mapping
Functions can be illustrated in three ways:
1) by a formula.
2) by arrow diagram.
3) by a graph (ie co-ordinate diagram).
ExampleSuppose that f: A B is defined by
f(x) = x2 + 3x where A = { -3, -2, -1, 0, 1}.FORMULA
then f(-3) = 0 , f(-2) = -2 , f(-1) = -2 , f(0) = 0 ,f(1) = 4
NB: B = {-2, 0, 4} = the range!
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0
-3
-2
-1
0
1
0
-2
-2
0
4
A B
ARROW DIAGRAM
Functions & MappingFunctions & Mapping
f(-3) = 0
f(-2) = -2
f(-1) = -2
f(0) = 0
f(1) = 4
f(x)
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Functions & GraphsFunctions & Graphs
In a GRAPH we get :
NB: This graph consists of 5 separate points. It is not a solid curve.
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Recognising FunctionsA B
a b c d
e
f
g
Not a function
two arrows leaving b!
A B
a bc d
e
f
g
YES
Functions & GraphsFunctions & Graphs
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Functions & GraphsFunctions & Graphs
A B
a b c d
e f g
Not a function - d unused!
A B
a bc d
e f g h
YES
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Higher
Functions & GraphsFunctions & Graphs
Recognising Functions from Graphs
If we have a function f: R R (R - real nos.) then every vertical line we could draw would cut
the graph exactly once!
This basically means that every x-value has one, and only one, corresponding y-value!
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Function & Graphs Function & Graphs
x
YFunction !
!
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x
YNot a
function !!
Cuts graph
more than once !
Function & GraphsFunction & Graphs
x must map to
one value of y
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Higher
Functions & GraphsFunctions & Graphs
X
Y Not a function !!
Cuts graph
more than once!
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X
YFunction !
!
Functions & GraphsFunctions & Graphs
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Nat 5
The standard way to represent a function
is by a formula.
Function Notation
Examplef(x) = x + 4
We read this as “f of x equals x + 4”
or
“the function of x is x + 4
f(1) = 5 is the value of f at 1
f(a) = a + 4 is the value of f at a
1 + 4 = 5
a + 4
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Nat 5
For the function h(x) = 10 – x2.
Calculate h(1) , h(-3) and h(5)
h(1) =
Examples
h(-3) = h(5) =
h(x) = 10 – x2
Function Notation
10 – 12 = 9
10 – (-3)2 =
10 – 9 = 1
10 – 52 =
10 – 25 = -15
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Nat 5
For the function g(x) = x2 + x
Calculate g(0) , g(3) and g(2a)
g(0) =
Examples
g(3) = g(2a) =
g(x) = x2 + x
Function Notation
02 + 0 =
0
32 + 3 =
12
(2a)2 +2a =
4a2 + 2a
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Nat 5 Outcome 1
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Inverse FunctionsInverse Functions
A Inverse function is simply a function in reverse
Input
Function
Outputf(x) = x2+ 3x - 1
InputOutputf-1(x) = ?
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Nat 5
Inverse Function
Find the inverse function given
f(x) = 3x
Example
Remember
f(x) is simply the
y-coordinate
y = 3x
Using Changing the subject
rearrange into
x =
x =y
3
Rewrite replacing y with
x.
This is the inverse function
f-1(x) =x
3
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Nat 5
Inverse Function
Find the inverse function given
f(x) = x2
Example
Remember
f(x) is simply the
y-coordinate
y = x2
Using Changing the subject
rearrange into
x =
x = √y
Rewrite replacing y with
x.
This is the inverse function
f-1(x) = √x
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Nat 5
Inverse Function
Find the inverse function given
f(x) = 4x - 1
Example
Remember
f(x) is simply the
y-coordinate
y = 4x - 1
Using Changing the subject
rearrange into
x =
x =
Rewrite replacing y with
x.
This is the inverse function
f-1(x) =
y + 1
4
x + 1
4
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COMPOSITION OF FUNCTIONS
( or functions of functions )
Suppose that f and g are functions where
f:A B and g:B C
with f(x) = y and g(y) = z
where x A, y B and z C.
Suppose that h is a third function where
h:A C with h(x) = z .
Composite FunctionsComposite Functions
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Composite FunctionsComposite Functions
A B C
x y zf g
h
We can say that h(x) = g(f(x))
“function of a function”
DEMO
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Composite FunctionsComposite Functionsf(2)=3x2 – 2 =4
g(4)=42 + 1 =17
f(5)=5x3-2 =13Example 1
Suppose that f(x) = 3x - 2 and g(x) = x2 +1
(a) g( f(2) ) = g(4) = 17
(b) f( g (2) ) = f(5) = 13
(c) f( f(1) ) = f(1) = 1
(d) g( g(5) ) = g(26)= 677
f(1)=3x1 - 2 =1
g(26)=262
+ 1 =677
g(2)=22 + 1 =5
f(1)=3x1 - 2 =1
g(5)=52 + 1 =26
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Suppose that f(x) = 3x - 2 and g(x) = x2 +1
Find formulae for (a) g(f(x)) (b) f(g(x)).
(a) g(f(x)) = ( )2 + 1
= 9x2 - 12x + 5
(b) f(g(x)) = 3( ) - 2= 3x2 + 1
CHECK
g(f(2)) = 9 x 22 - 12 x 2 + 5
= 36 - 24 + 5= 17
f(g(2)) = 3 x 22 + 1= 13
NB: g(f(x)) f(g(x)) in general.
Composite FunctionsComposite Functions
3x - 2 x2 +1
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Let h(x) = x - 3 , g(x) = x2 + 4 and k(x) = g(h(x)). If k(x) = 8 then find the value(s) of x.
k(x) = g(h(x))
= ( )2 + 4
= x2 - 6x + 13
Put x2 - 6x + 13 = 8
then x2 - 6x + 5 = 0
or (x - 5)(x - 1) = 0
So x = 1 or x = 5
Composite FunctionsComposite Functions
x - 3
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Choosing a Suitable Domain
(i) Suppose f(x) = 1 . x2 - 4
Clearly x2 - 4 0
So x2 4
So x -2 or 2
Hence domain = {xR: x -2 or 2 }
Composite FunctionsComposite Functions
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(ii) Suppose that g(x) = (x2 + 2x - 8)
We need (x2 + 2x - 8) 0
Suppose (x2 + 2x - 8) = 0
Then (x + 4)(x - 2) = 0
So x = -4 or x = 2
So domain = { xR: x -4 or x 2 }
Composite FunctionsComposite FunctionsSketch graph
-4 2
Graphs & Functions Higher
The functions f and g are defined on a suitable domain by
a) Find an expression for b) Factorise
2 2( ) 1 and ( ) 2f x x g x x
( ( ))f g x ( ( ))f g x
a) 22 2( ( )) ( 2) 2 1f g x f x x
2 22 1 2 1x x Difference of 2 squares
Simplify 2 23 1x x
b)
Graphs & Functions Higher
Functions and are defined on suitable domains.
a) Find an expression for h(x) where h(x) = f(g(x)).
b) Write down any restrictions on the domain of h.
1( )
4f x
x
( ) 2 3g x x
( ( )) (2 3)f g x f x a)1
2 3 4x
1
( )2 1
h xx
b) 2 1 0x 1
2x
Graphs & Functions Higher
3( ) 3 ( ) , 0x
f x x and g x x
a) Find
b) If find in its simplest form.
( ) where ( ) ( ( ))p x p x f g x
3
3( ) , 3
xq x x
( ( ))p q x
3( ) ( ( ))x
p x f g x f
a) 33
x 3 3x
x
3( 1)x
x
b)
33 1
3333
3
( ( )) x
xx
p q x p
9 3
33 3x x
9 3(3 ) 3
3 3
x x
x
3 3
3 3
x x
x
x
Graphs & Functions HigherFunctions f and g are defined on the set of real numbers by
a) Find formulae for
i) ii)
b) The function h is defined by
Show that and sketch the graph of h.
2( ) 1 and ( )f x x g x x
( ( ))f g x ( ( ))g f x
( ) ( ( )) ( ( ))h x f g x g f x
2( ) 2 2h x x x
a)
b)
2 2( ( )) ( ) 1f g x f x x 2( ( )) ( 1) 1g f x g x x
22( ) 1 1h x x x 2 2( ) 1 2 1h x x x x 22 2x x
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A function in the form f(x) = ax where a > 0, a ≠ 1
is called an exponential function to base a .
Exponential (to the power of) Graphs
Exponential Functions
Consider f(x) = 2x
x -3 -2 -1 0 1 2 3
f(x) 1 1/8 ¼ ½ 1 2 4 8
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The graph of
y = 2x
(0,1)(1,2)
Major Points
(i) y = 2x passes through the points (0,1) & (1,2)
(ii) As x ∞ y ∞ however as x -∞ y 0 .(iii) The graph shows a GROWTH function.
Graph
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ie
y -3 -2 -1 0 1 2 3
x 1/8 ¼ ½ 1 2 4 8
To obtain y from x we must ask the question
“What power of 2 gives us…?”
This is not practical to write in a formula so we say
y = log2x“the logarithm to base 2 of x”
or “log base 2 of x”
Log Graphs
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The graph of
y = log2x (1,0)
(2,1)
Major Points
(i) y = log2x passes through the points (1,0) & (2,1) .(ii) As x ∞ y ∞ but at a very slow rate
and as x 0 y -∞ .
NB: x > 0
Graph
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The graph of y = ax always passes through (0,1) & (1,a)
It looks like ..
x
Y
y = ax
(0,1)
(1,a)
Exponential (to the power of) Graphs
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The graph of y = logax always passes through (1,0) & (a,1)
It looks like ..
x
Y
y = logax
(1,0)
(a,1)
Log Graphs
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Graph Transformations
We will investigate f(x) graphs of the form
1. f(x) ± k
2. f(x ± k)
3. -f(x)
4. f(-x)
5. kf(x)
6. f(kx)
Each moves the
Graph of f(x) in a certain
way !
f(x)
0 2 4 6 8x
-2-4-6
2
4
6
-2
-4
-6
Transformation f(x) ± k
(x , y) (x , y ± k)
Mapping
f(x) + 5 f(x) - 3
f(x)
Transformation f(x) ± k
Keypoints
y = f(x) ± k
moves original f(x) graph vertically up or down
+ k move up
- k move down
Only y-coordinate changes
NOTE: Always state any coordinates given on f(x)
on f(x) ± k graph
Demo
f(x) - 2
A(-1,-2) B(1,-2)C(0,-3)
f(x) + 1
B(90o,0)
A(45o,0.5)
C(135o,-0.5)
B(90o,1)
A(45o,1.5)
C(135o,0.5)
f(x)
0 2 4 6 8x
-2-4-6
2
4
6
-2
-4
-6
Transformation f(x ± k)
(x, y) (x ± k , y)
Mapping
f(x - 2) f(x + 4)
f(x)
Transformation f(x ± k)
Keypoints
y = f(x ± k)
moves original f(x) graph horizontally left or right
+ k move left
- k move right
Only x-coordinate changes
NOTE: Always state any coordinates given on f(x)
on f(x ± k) graph
Demo
f(x)
0 2 4 6 8x
-2-4-6
2
4
6
-2
-4
-6
Transformation -f(x)
(x, y) (x , -y)
Mapping
f(x)
Flip inx-axis
Flip inx-axis
Transformation -f(x)
Keypoints
y = -f(x)
Flips original f(x) graph in the x-axis
y-coordinate changes sign
NOTE: Always state any coordinates given on f(x)
on -f(x) graph
Demo
- f(x)
A(-1,0) B(1,0)
C(0,1)
- f(x)
B(90o,0)
A(45o,0.5)
C(135o,-0.5)A(45o,-0.5)
C(135o,0.5)
f(x)
0 2 4 6 8x
-2-4-6
2
4
6
-2
-4
-6
Transformation f(-x)
(x, y) (-x , y)
Mapping
f(x)
Flip iny-axis
Flip iny-axis
Transformation f(-x)
Keypoints
y = f(-x)
Flips original f(x) graph in the y-axis
x-coordinate changes sign
NOTE: Always state any coordinates given on f(x)
on f(-x) graph
Demo
f(-x)
B(0,0)
C’(-1,1)
A’(1,-1)A(-1,-1)
C (1,1)
f(x)
0 2 4 6 8x
-2-4-6
2
4
6
-2
-4
-6
Transformation kf(x)
(x, y) (x , ky)
Mapping
f(x)
Stretch iny-axis
2f(x) 0.5f(x)
Compress iny-axis
Transformation kf(x)
Keypoints
y = kf(x)
Stretch / Compress original f(x) graph in the
y-axis direction
y-coordinate changes by a factor of k
NOTE: Always state any coordinates given on f(x)
on kf(x) graph
Demo
f(x)
0 2 4 6 8x
-2-4-6
2
4
6
-2
-4
-6
Transformation f(kx)
(x, y) (1/kx , y)
Mapping
f(x)
Compress inx-axis
f(2x) f(0.5x)
Stretch inx-axis
Transformation f(kx)
Keypoints
y = f(kx)
Stretch / Compress original f(x) graph in the
x-axis direction
x-coordinate changes by a factor of 1/k
NOTE: Always state any coordinates given on f(x)
on f(kx) graph
Demo
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You need to be able to work with combinations
Combining Transformations
Demo
(1,3)
(-1,-3)
(1,3)
(-1,-3)
2f(x) + 1
f(0.5x) - 1
f(-x) + 1
-f(x + 1) - 3
Explain the effect the following have
(a)-f(x)
(b)f(-x)
(c)f(x) ± k
Explain the effect the following have
(d)f(x ± k)
(e)kf(x)
(f)f(kx)
Name :
(-1,-3)
(1,3)
(-1,-3)
(1,3)
f(x + 1) + 2
-f(x) - 2
(1,3)
(-1,-3)
(-1,-3)
(1,3)
(1,3)
(-1,-3)
(1,-2)
(1,3)(-1,4)
(-1,-3)(1,-5)
(1,3)
(-1,1)
(-1,-3)
(0,5)
2f(x) + 1
f(0.5x) - 1
f(-x) + 1
-f(x + 1) - 3
Explain the effect the following have
(a)-f(x) flip in x-axis
(b)f(-x) flip in y-axis
(c)f(x) ± k move up or down
Explain the effect the following have
(d)f(x ± k) move left or right
(e)kf(x) stretch / compressin y
direction
(e)f(kx) stretch / compress
in x direction
Name :
(-1,-3)
(1,3)
(-1,-3)
(1,3)
(-2,-1)
f(x + 1) + 2
-f(x) - 2
(-2,0)
(1,3)
(0,-6)
(-1,-3)
(-1,-3)
(1,3)
(1,7)
(-1,-5)
(2,2)
(1,3)
(-2,-4)(-1,-3)
The diagram shows the graph of a function f.
f has a minimum turning point at (0, -3) and a
point of inflexion at (-4, 2).
a) sketch the graph of y = f(-x).
b) On the same diagram, sketch the graph of y = 2f(-x)
Graphs & Functions Higher
a) Reflect across the y axis
b) Now scale by 2 in the y direction-1 3 4
2
y = f(-x)
-3
y
x
4
y = 2f(-x)
-6
Graphs & Functions Higher
Part of the graph of is shown in the diagram.
On separate diagrams sketch the graph of
a) b)
Indicate on each graph the images of O, A, B, C, and D.
( )y f x
( 1)y f x 2 ( )y f x
a)
b)
graph moves to the left 1 unit
graph is reflected in the x axis
graph is then scaled 2 units in the y direction
(2, 1)
(2, -1)
(2, 1)
5
y=f(x)
y= -f(x)
y= 10 - f(x)
Graphs & Functions Higher =
a) On the same diagram sketch
i) the graph of
ii) the graph of
b) Find the range of values of x for
which is positive
2( ) 4 5f x x x
( )y f x
10 ( )y f x
10 ( )f x
2( 2) 1x
a)
b) Solve:210 ( 2) 1 0x
2( 2) 9x ( 2) 3x 1 or 5x
10 - f(x) is positive for -1 < x < 5
Graphs & Functions Higher
A sketch of the graph of y = f(x) where is shown.
The graph has a maximum at A (1,4) and a minimum at B(3, 0)
.
Sketch the graph of
Indicate the co-ordinates of the turning points. There is no need to
calculate the co-ordinates of the points of intersection with the axes.
3 2( ) 6 9f x x x x
( ) ( 2) 4g x f x
Graph is moved 2 units to the left, and 4 units up(3, 0) (1, 4)
(1, 4) ( 1, 8)
t.p.’s are:
(1,4)
(-1,8)
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Trig Graphs
The same transformation rules
apply to the basic trig graphs.
NB: If f(x) =sinx then 3f(x) = 3sinx
and f(5x) = sin5x
Think about sin replacing f !
Also if g(x) = cosx then g(x) – 4 = cosx – 4
and g(x + 90) = cos(x + 90) Think about cos replacing g !
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Sketch the graph of y = sinx - 2
If sinx = f(x) then sinx - 2 = f(x) - 2
So move the sinx graph 2 units down.
y = sinx - 2
Trig Graphs
1
-1
-2
-3
090o 180o 270o 360o
DEMO
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Sketch the graph of y = cos(x - 50)
If cosx = f(x) then cos(x - 50) = f(x - 50)So move the cosx graph 50 units right.
Trig Graphs
y = cos(x - 50)o
1
-1
-2
-3
0
50o
90o 180o 270o 360o
DEMO
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Trig Graphs
Sketch the graph of y = 3sinx
If sinx = f(x) then 3sinx = 3f(x)
So stretch the sinx graph 3 times vertically.
y = 3sinx
1
-1
-2
-3
0
2
3
90o 180o 270o 360o
DEMO
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Trig Graphs
Sketch the graph of y = cos4x
If cosx = f(x) then cos4x = f(4x)
So squash the cosx graph to 1/4 size horizontally
y = cos4x
1
-1
090o 180o 270o 360o
DEMO
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Trig Graphs
Sketch the graph of y = 2sin3xIf sinx = f(x) then 2sin3x = 2f(3x)So squash the sinx graph to 1/3 size horizontally and also double its height.
y = 2sin3x
90o
1
-1
-2
-3
0
2
3
360o180o 270o
DEMO
DEMO
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Trig Graph Trig Graph
1
2
3
-3
-2
-1
090o 180o 270o 360o
Write down equations for
graphs shown ?
CombinationsHigher
y = 0.5sin2xo + 0.5
y = 2sin4xo- 1
Write down the equations in the form f(x) for the graphs shown?
y = 0.5f(2x) + 0.5
y = 2f(4x) - 1
DEMO
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Trig GraphsTrig Graphs
1
2
3
-3
-2
-1
090o 180o 270o 360o
Combinationsy = cos2xo + 1
y = -2cos2xo - 1Higher
Write down the equations for the graphs shown?
Write down the equations in the form f(x) for the graphs shown?
y = f(2x) + 1y = -2f(2x) - 1
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Radians
Radian measure is an alternative to degrees and is based upon the ratio of
arc Length radius
rθ
L
θ- theta
(angle at the centre)
Circumf erence 2 r
2 Circumference
r
So, full circle 360o 2π radians
RadiansCopy Table360o 2π180o π
90o π2
45o π4
30o π6
60o π3
270o 3π2
135o 3π4
150o 5π6
120o 2π3
225o 5π4
210o 7π6
240o 4π3
315o 7π4
330o
300o 5π3
11π6
Demo
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Converting
degrees radians
÷180then X π
÷ π then x
180
For any values
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Ex1 72o =72/180 X π = 2π /5
Ex2 330o =330/180 X π =11 π /6
Ex3 2π /9 =2π /9 ÷ π x 180o = 2/9 X 180o = 40o
Ex4 23π/18 = 23π /18 ÷ π x 180o
= 23/18 X 180o = 230o
Converting
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22
2
60º
60º60º160º
230º3
This triangle will provide exact values for
sin, cos and tan 30º and 60º
Exact Values
Some special values of Sin, Cos and Tan are useful left as fractions, We call these exact values
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x 0º 30º 45º 60º 90º
Sin xº
Cos xº
Tan xº
½
½
3
3
2
3
20
1
0
1
0
Exact Values
1
3
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Exact Values
1 1 45º
45º
2
Consider the square with sides 1 unit
11
We are now in a position to calculate exact values for sin, cos and tan of 45o
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x 0º 30º 45º 60º 90º
Sin xº
Cos xº
Tan xº Undefined
½
½
3
3
2
3
20
1
0
1
0
Exact Values
1
3
1 2
1 2
1
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Exact value table
and quadrant rules.
tan150o
(Q2 so neg)
= - tan(180 - 150) o
= - tan30o= -1/√3
cos300o
(Q4 so pos)
= cos(360 - 300) o
= cos60o
= 1/2
sin120o
(Q2 so pos)
= sin(180 - 120)
o
= sin60o= √ 3/2
tan300o
(Q4 so neg)
= - tan(360-300)o
= - tan60o= - √ 3
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Find the exact value of cos2(5π/6) – sin2(π/6)
cos(5π/6) =
cos150o
(Q2 so neg)
= cos(180 - 150)o
= - cos30o= - √3 /2
sin(π/6) = sin30o = 1/2
cos2(5π/6) – sin2(π/6) = (- √3 /2)2 – (1/2)2= ¾ - 1/4 = 1/2
Exact value table
and quadrant rules.
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Exact value table
and quadrant rules.
Prove that sin(2 π /3) = tan (2 π /3)
cos (2 π /3) sin(2π/3) = sin120o = sin(180 – 120)o=
sin60o= √3/2
cos(2 π /3) = cos120o
tan(2 π /3) = tan120o
= cos(180 – 120)o
= tan(180 – 120)o
= - cos60o
= -tan60o
= -1/2
= - √3
LHS =sin(2 π /3) cos (2 π /3)
= √3/2 ÷ -1/2 = √3/2 X -2
= - √3 = tan(2π/3) = RHS
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created by Mr. Laffertycreated by Mr. Lafferty
Solving Trig Equations Solving Trig Equations
All +veSin +ve
Tan +ve Cos +ve
180o - xo
180o + xo 360o - xo
1 2 3 4
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created by Mr. Laffertycreated by Mr. Lafferty
Solving Trig Equations
Example 1 Type 1:
Solving the equation sin xo = 0.5 in the range 0o to 360o
Graphically what are we
trying to solve
xo = sin-1(0.5)
xo = 30o
There is another solution
xo = 150o
(180o – 30o = 150o)
sin xo = (0.5)
1 2 3 4
C
AS
T0o180
o
270o
90o
3
2
2
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created by Mr. Laffertycreated by Mr. Lafferty
Solving Trig Equations
Example 2 :
Solving the equation cos xo - 0.625 = 0 in the range 0o to 360o
Graphically what are we
trying to solve
cos xo = 0.625
xo = 51.3o
(360o - 53.1o = 308.7o)
xo = cos -1 (0.625)
There is another solution
1 2 3 4xo = 308.7o
C
AS
T0o180
o
270o
90o
3
2
2
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created by Mr. Laffertycreated by Mr. Lafferty
Solving Trig Equations
Example 3 :
Solving the equation tan xo – 2 = 0 in the range 0o to 360o
Graphically what are we
trying to solve
tan xo = 2
xo = 63.4o
x = 180o + 63.4o = 243.4o
xo = tan -1(2)
There is another solution
1 2 3 4
C
AS
T0o180
o
270o
90o
3
2
2
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created by Mr. Laffertycreated by Mr. Lafferty
Solving Trig Equations
Example 4 Type 2 :
Solving the equation sin 2xo + 0.6 = 0 in the range 0o to 360o
Graphically what are we
trying to solve
2xo = sin-1(0.6)
2xo = 217o , 323o
577o , 683o ......
sin 2xo = (-0.6)
xo = 108.5o , 161.5o
288.5o , 341.5o
C
AS
T0o180
o
270o
90o
3
2
2
2xo = 37o ( always 1st Q First)
÷2
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created by Mr. Laffertycreated by Mr. Lafferty
Solving Trig EquationsGraphically what are we
trying to solve
sin (2x - 30o) = √3 ÷ 2
2xo - 30o = 60o , 120o ,420o , 480o .........
2sin (2x - 30o) = √3
xo = 45o , 75o
225o , 255o
2x - 30o = sin-1(√3 ÷ 2)
Example 5 Type 3 :
Solving the equation 2sin (2xo - 30o) - √3 = 0 in the range 0o to 360o
C
AS
T0o180
o
270o
90o
3
2
2
2xo = 90o , 150o ,450o , 510o .........
÷2
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created by Mr. Laffertycreated by Mr. Lafferty
Solving Trig Equations
Example 6 Type 4 :
Solving the equation cos2x = 1 in the range 0o to 360o
Graphically what are we
trying to solve
cos xo = ± 1
cos xo = 1
cos2 xo = 1
xo = 0o and 360o
C
AS
T0o180
o
270o
90o
3
2
2
cos xo = -1xo = 180o
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Example 7 Type 5 : Solving the equation 3sin2x + 2sin x - 1 = 0 in the range 0o to 360o
Solving Trig Equations
3p – 1 = 0
xo = 19.5o and 160.5o
xo = 270o
Let p = sin x
We have 3p2 + 2p - 1 = 0
(3p – 1)(p + 1) = 0Factorise
p = 1/3
p + 1 = 0p = - 1sin x =
1/3sin x = -1
C
AS
T0o180
o
270o
90o
3
2
2
C
AS
T0o180
o
270o
90o
3
2
2
Graphs & Functions Higher
Functions f and g are defined on suitable domains by and
a) Find expressions for:
i)
ii)
b) Solve
( ) sin( )f x x ( ) 2g x x
( ( ))f g x
( ( ))g f x
2 ( ( )) ( ( )) 0 360f g x g f x for x
( ( )) (2 )f g x f xa) sin 2x ( ( )) (sin )g f x g x 2sin x
b) 2sin 2 2sinx x sin 2 sin 0x x
2sin cos sin 0x x x sin (2cos 1) 0x x 1
or2
sin 0 cosx x 0 , 180 , 360x 60 , 300x
Graphs & Functions Higher
Functions
are defined on a suitable set of real numbers.
a) Find expressions for
b) i) Show that
ii) Find a similar expression for
and hence solve the equation
4and( ) sin , ( ) cos ( )f x x g x x h x x
( ( ))f h x ( ( ))g h x
1 1
2 2( ( )) sin cos f h x x x
( ( ))g h x
for( ( )) ( ( )) 1 0 2f h x g h x x
4( ( )) ( )f h x f x a) 4
sin( )x 4
( ( )) cos( )g h x x
sin cos4 4 4
sin( ) sin cos xx x b) Now use exact values
Repeat for ii)
equation reduces to2
sin 12
x 2 1sin
2 2x
3,
4 4x
Graphs & Functions Higher
The diagram shows a sketch of part of
the graph of a trigonometric function
whose equation is of the form
Determine the values of a, b and c
sin( )y a bx c
a is the amplitude: a = 4
b is the number of waves in 2 b = 2
c is where the wave is centred vertically c = 1
2a
1 in
2 in 2
1
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An identity is a statement which is true for all values.
eg 3x(x + 4) = 3x2 + 12x
eg (a + b)(a – b) = a2 – b2
Trig Identities
(1) sin2θ + cos2 θ = 1
(2) sin θ = tan θ cos θ θ ≠ an odd multiple of π/2 or 90°.
Trig Identities
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Reason
θo
c
b
a
a2 +b2 = c2
sinθo = a/c
cosθo = b/c
(1) sin2θo + cos2 θo =
Trig Identities
2 2
2 2
a bc c
2 2
2
a bc
2
2 1cc
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sin2θ + cos2 θ = 1
sin2 θ = 1 - cos2
θ cos2 θ = 1 - sin2
θ
Simply rearranging we get two other forms
Trig Identities
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Example1 sin θ = 5/13 where 0 < θ < π/2
Find the exact values of cos θ and tan θ .
cos2 θ = 1 - sin2 θ
= 1 – (5/13)2
= 1 – 25/169= 144/169
cos θ = √(144/169)
= 12/13 or -12/13
Since θ is between 0 < θ < π/2
then cos θ > 0
So cos θ = 12/13
tan θ = sinθ cos θ
= 5/13 ÷ 12/13
= 5/13 X 13/12
tan θ = 5/12
Trig Identities
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Given that cos θ = -2/ √ 5 where π< θ < 3 π /2Find sin θ and tan θ.
sin2 θ = 1 - cos2 θ
= 1 – (-2/ √
5 )2
= 1 – 4/5
= 1/5
sin θ = √(1/5)
= 1/ √ 5 or - 1/ √ 5
Since θ is between π< θ < 3 π /2
sinθ < 0
Hence sinθ = - 1/√5
tan θ = sinθ cos θ
= - 1/ √ 5 ÷ -2/ √
5 = - 1/ √ 5 X - √5 /2
Hence tan θ = 1/2
Trig Identities
Outcome 2
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Higher
Are you on Target !
• Update you log book
• Make sure you complete and correct MOST of the Composite Functionquestions in the past paper booklet.
• Make sure you complete and correct SOME of the Trigonometryquestions in the past paper booklet.
f(x)
f(x)
Graphs & Functions
y = -f(x)
y = f(-x)
y = f(x) ± k
y = f(kx)
Move verticallyup or downs
depending on k
flip iny-axis
flip inx-axis
+
- Stretch or compressvertically
depending on k
y = kf(x)
Stretch or compress
horizontally depending on k
f(x)
f(x)f(x)
f(x)y = f(x ± k)
Move horizontallyleft or right
depending on k
+-
Remember we can combine
these together !!
0 < k < 1 stretch
k > 1 compress
0 < k < 1 compress
k > 1 stretch
Composite Functions
A complex function made up of 2 or
more simpler functions
= +
f(x) = x2 - 4 g(x) = 1x
x
Domainx-axis valuesInput
Rangey-axis valuesOutput
x2 - 41
x2 - 4
Restriction x2 - 4 ≠ 0
(x – 2)(x + 2) ≠ 0
x ≠ 2 x ≠ -2
g(f(x)) g(f(x)) =
f(x) = x2 - 4g(x) = 1x
x
Domainx-axis valuesInput
Rangey-axis valuesOutput
f(g(x))
Restriction x2 ≠ 0
1
x
2- 4 =
Similar to composite
Area
Write down g(x) with brackets for x
g(x) =1
( )
inside bracket put f(x)
g(f(x)) =1
x2 - 4
1x
- 41x2
f(g(x)) =Write down f(x) with brackets for x
f(x) = ( )2 - 4
inside bracket put g(x)
f(g(x)) =1
x2- 4
Functions & Graphs
TYPE questions(Sometimes Quadratics)
SketchingGraphs
CompositeFunctions
Steps :
1.Outside function staysthe same EXCEPT replacex terms with a ( )
2. Put inner function in bracket
You need to learn basic movements
Exam questionsnormally involve two movements
Remember orderBODMAS
Restrictions :
1.Denominator NOT ALLOWEDto be zero
2.CANNOT take the square rootof a negative number